# Covariant derivative of determinant of the metric tensor

Let $$(M,g)$$ be a Riemannian manifold and $$g$$ the Riemannian metric in coordinates $$g=g_{\alpha \beta}dx^{\alpha} \otimes dx^{\beta}$$, where $$x^{i}$$ are local coordinates on $$M$$. Denote by $$g^{\alpha \beta}$$ the inverse components of the inverse metric $$g^{-1}$$. Let $$\nabla$$ be the Levi-Civita connection of the metric $$g$$. Consider, locally, the function $$\det((g_{\alpha \beta})_{\alpha \beta})$$. It is known that $$\nabla \det((g_{\alpha \beta})_{\alpha \beta}) = 0$$ by using normal coordinates etc...

I would like to show this fact without using normal coordinates. Just by computation. Here is what I have so far:

$$\nabla \det((g_{\alpha \beta})_{\alpha \beta}) = \left [ g^{\gamma \delta} \partial_{\delta} \det((g_{\alpha \beta})_{\alpha \beta}) \right ] \partial_{\gamma} = \left [ \det((g_{\alpha \beta})_{\alpha \beta}) g^{\gamma \delta} g^{\beta \alpha} \partial_{\delta} g_{\alpha \beta}\right ] \partial_{\gamma}.$$

Here: the first equality sign follows from the definition of the gradient of a function and the second equality sign is the derivative of the determinant.

Question: How do I continue from here without using normal coordinates? Or are there any mistakes? If yes, where and which?

• Is $\det g$ really a well-defined function on $M$? – Arctic Char Apr 14 at 10:43
• It is not defined on the whole of $M$, only in a chart. – Phillip Apr 14 at 11:10
• Then I doubt if it makes any sense to calculate this using normal coordinates. The function is not only defined in a chart, but only depends on the chart chosen. So when calculate the gradient you cannot just switch to normal coordinates (which is another chart) – Arctic Char Apr 14 at 11:14
• I meant to say the claim that the gradient of det g is zero is false. You are requiring the function to be constant, but you have no restriction on g whatsoever. $g= x dx^2$ defined on some open interval of $\mathbb R$ will be a counterexample. – Arctic Char Apr 14 at 11:28
• It seems like you are confusing covariant derivative with gradient. These are two different concepts, even though both are denoted by $\nabla$. – Amitai Yuval Apr 14 at 12:51

Calculate $$\Gamma^\nu_{\mu\nu}=\frac{1}{2}g^{\nu\kappa}(\partial_\mu g_{\nu\kappa}+\partial_\nu g_{\mu\kappa}-\partial_\kappa g_{\mu\nu})=\frac{1}{2}g^{\nu\kappa}\partial_\mu g_{\nu\kappa}$$. Using the well known formula $$\det A^{-1}\frac{d}{dt}\det A=\text{Tr}\left(A^{-1}\frac{d A}{dt}\right),$$ we obtain $$\Gamma_\mu\equiv\Gamma^\nu_{\mu\nu}=\frac{1}{2}\frac{1}{g}\partial_\mu g=\frac{1}{2}\partial_\mu\ln|g|=\partial_\mu\ln \sqrt{|g|},$$ where $$g$$ denotes the determinant of the matrix with elements $$g_{\mu\nu}$$.
The expression $$\rho=\sqrt{|g|}$$ transforms under a change of chart as follows: We have $$g_{\mu^\prime \nu^\prime}=\partial_{\mu^\prime}x^\mu\partial_{\nu^\prime}x^\nu g_{\mu\nu}=J^\mu_{\mu^\prime}J^\nu_{\nu^\prime}g_{\mu\nu}$$, taking determinants gives $$g^\prime=\det J^2g \\ \sqrt{|g^\prime|}=|\det J|\sqrt{|g|}.$$ This object is called a "scalar density of weight 1". It makes sense in a coordinate-free manner too as a section of the density bundle, but whatever. One can show that the components of the covariant derivative of such an object is $$\nabla_\mu\rho=\partial_\mu\rho-\Gamma_\mu\rho=\partial_\mu\rho-\partial_\mu\ln\sqrt{|g|}\rho.$$ Inserting $$\rho=\sqrt{|g|}$$ gives $$\nabla_\mu\sqrt{|g|}=\partial_\mu\sqrt{|g|}-\frac{1}{\sqrt{|g|}}\partial_\mu\sqrt{|g|}\sqrt{|g|}=\partial_\mu\sqrt{|g|}-\partial_\mu\sqrt{|g|}=0.$$
One can give a short proof by considering the reciprocal basis $$\partial^k=g^{k\sigma}\partial_{\sigma}$$ and its derivative $$\nabla_{\partial_k}\partial^j=-\Gamma^{j}{}_{ks}\partial^s$$, because if we represent the metric tensor as $$g=g^{st}\partial_s\otimes\partial_t$$ then $$\begin{eqnarray*} g&=&g^{st}\partial_s\otimes\partial_t\\ &=&\partial_s\otimes g^{st}\partial_t\\ &=&\partial_s\otimes\partial^s. \end{eqnarray*}$$
So $$\begin{eqnarray*} \nabla_{\partial_k}g&=&\nabla_{\partial_k}(\partial_s\otimes\partial^s)\\ &=&(\nabla_{\partial_k}\partial_s)\otimes\partial^s+\partial_s\otimes\nabla_{\partial_k}\partial^s\\ &=&\Gamma^r{}_{ks}\partial_r\otimes\partial^s-\partial_s\otimes \Gamma^s{}_{kr}\partial^r\\ &=&\Gamma^r{}_{ks}\partial_r\otimes\partial^s-\Gamma^s{}_{kr}\partial_s\otimes \partial^r\\ &=&\Gamma^r{}_{ks}\partial_r\otimes\partial^s-\Gamma^r{}_{ks}\partial_r\otimes \partial^s\\ &=&0. \end{eqnarray*}$$